1
$\begingroup$

From previous chapters, I think I have understand the twist, it is about the angular velocity and the linear velocity, the difference is that linear velocity is not one fixed point in the body but is the one in the body in coincidence with the origin instantly.

I also notice the difference Stationary/inertial reference frame

first quetion

in chapter 8.2 ,is $v_b$ the velocity of the mass center of the body? after all the inertia frame is in coincidence with the mass center instantly.

Let $r_i =(x_i, y_i, z_i)$ be the fixed location of mass $i$ in a body frame {b}, where the origin of this frame is the unique point such that $\sum_{i} m_i r_i =0$. This point is known as the center of mass.

second question

what is the relationship between eq8.22 $f_b = m(\dot v_b + [\omega_b]v_b)$ and $f_b = m(\dot v_c)$ from classical method ,$v_c$ is the linear velocity of the mass center

third question

I have read the Roy Featherstone book, the spatial acceleration is defined as $[\dot w, \ddot r - w \times \dot r]^T$ , this garantees that spatial acceleration is zero even when the body is rotating at a constant speed therefore some points are undergoing centripetal force. However in the modern robotics I find that the derivative of the twist is not accordance with the the spatial acceleration

$\endgroup$
0
$\begingroup$

To answer your first question: $v_b$ is the linear velocity of the centre of mass of the body relative to a fixed inertial frame coincident with the body frame with its origin at the centre of mass.

For your second question: Equation 8.22 takes into account the contributions of all linear and rotary motions on the overall force on the body, while the equation you've included from the classical method ($f_b=m(\dot v_c)$) assumes you are both measuring in a fixed inertial frame and that the body is not rotating.

Finally, the third question: there are a few key differences in the notation used in the Modern Robotics textbook and in Featherstone's work, one of which produces equations for spatial acceleration that appear to disagree with each other.

However, having used both methods both in my research and in teaching courses on robotics for many years, I can assure you that as long as you stick with one standard of notation you will end up with the same numbers in the end. In other words, $\dot v_b + [\omega_b]v_b = \ddot r - \omega \times \dot r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.